Metamath Proof Explorer


Theorem cbvabv

Description: Rule used to change bound variables, using implicit substitution. Version of cbvab with disjoint variable conditions requiring fewer axioms. (Contributed by NM, 26-May-1999) Require x , y be disjoint to avoid ax-11 and ax-13 . (Revised by Steven Nguyen, 4-Dec-2022)

Ref Expression
Hypothesis cbvabv.1 ⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
Assertion cbvabv { 𝑥 ∣ 𝜑 } = { 𝑦 ∣ 𝜓 }

Proof

Step Hyp Ref Expression
1 cbvabv.1 ⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
2 1 cbvsbv ⊢ ( [ 𝑧 / 𝑥 ] 𝜑 ↔ [ 𝑧 / 𝑦 ] 𝜓 )
3 df-clab ⊢ ( 𝑧 ∈ { 𝑥 ∣ 𝜑 } ↔ [ 𝑧 / 𝑥 ] 𝜑 )
4 df-clab ⊢ ( 𝑧 ∈ { 𝑦 ∣ 𝜓 } ↔ [ 𝑧 / 𝑦 ] 𝜓 )
5 2 3 4 3bitr4i ⊢ ( 𝑧 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝑧 ∈ { 𝑦 ∣ 𝜓 } )
6 5 eqriv ⊢ { 𝑥 ∣ 𝜑 } = { 𝑦 ∣ 𝜓 }